The Poisson equation with $L^2$ data has a unique weak solution in $H^1$

Question

In picture below, I have proved that if $f\in L^2(U)$ , (*) has unique weak solution. I use the Lax Milgran to prove it. I think some conditions are redundant.

Answer 1

No, there are no redundant conditions here.

1. The equation $\int_U Du\cdot Dv = \int_U fv$ is not a new condition: it is the definition of what it means for $u$ to be a weak solution of $-\Delta u = f$. In other words, it replaces $-\Delta u=f$, which does not hold literally since $u$ need not be twice differentiable.

2. If $\int_U f\ne 0$, then there is no solution. Just put $v\equiv 1$ in the definition of the weak solution, above, and $0=\int_U f$ follows.

By the way, the weak solution is not unique. There are infinitely many, since one can add a constant without affecting the Neumann condition.

And if you are interested in the actual solution, see Weak solutions to the Neumann's problem (Evans PDE)